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A192722 T(n,k) = Sum of multinomial(n; n_1,n_2,...,n_k)^2, where the sum extends over all compositions (n_1,n_2,...,n_k) of n into exactly k nonnegative parts.

%I A192722 #29 Feb 15 2025 05:36:22
%S A192722 1,1,4,1,18,36,1,68,432,576,1,250,3900,14400,14400,1,922,32400,252000,
%T A192722 648000,518400,1,3430,262542,3880800,19404000,38102400,25401600,1,
%U A192722 12868,2119152,56664384,493920000,1795046400,2844979200,1625702400
%N A192722 T(n,k) = Sum of multinomial(n; n_1,n_2,...,n_k)^2, where the sum extends over all compositions (n_1,n_2,...,n_k) of n into exactly k nonnegative parts.
%C A192722 Compare with triangle A019538, whose entries are given by
%C A192722 ... Sum multinomial(n; n_1,n_2,...,n_k), where the sum extends over all compositions (n_1,n_2,...,n_k) of n into exactly k nonnegative parts.
%C A192722 For related tables see A061691 and A192721.
%C A192722 Let P be the poset of all ordered pairs (S,T) of subsets of [n] with |S|=|T|, ordered componentwise by inclusion.  T(n,k) is the number of length k chains in P from ({},{}) to ([n],[n]). - _Geoffrey Critzer_, Apr 15 2020
%H A192722 Alois P. Heinz, <a href="/A192722/b192722.txt">Rows n = 1..100, flattened</a>
%F A192722 Generating function: Let J(z) = Sum_{n>=0} z^n/n!^2. Then
%F A192722 1 + Sum_{n>=1} (Sum_{k = 1..n} T(n,k)*x^k)*z^n/n!^2 = 1/(1 - x*(J(z) - 1))
%F A192722   = 1 + x*z + (x + 4*x^2)*z^2/2!^2 + (x + 18*x^2 + 36*x^3)*z^3/3!^2 + ....
%F A192722 Relations with other sequences:
%F A192722 The change of variable z -> z/x followed by x -> 1/(x - 1) transforms the above bivariate generating function 1/(1 - x*(J(z) - 1)) into (1 - x)/(-x + J(z*(x-1))), which is the generating function for A192721.
%F A192722 1/k!*T(n,k) = A061691(n,k).
%F A192722 T(n,n) = n!^2 = A001044(n).
%F A192722 Row sums = A102221.
%F A192722 For n>=1, Sum_{k = 1..n} (-1)^(n+k)*T(n,k)/k = A002190(n).
%e A192722 The triangle begins
%e A192722 n/k|..1.....2.......3........4........5........6
%e A192722 ================================================
%e A192722 .1.|..1
%e A192722 .2.|..1.....4
%e A192722 .3.|..1....18.....36
%e A192722 .4.|..1....68.....432......576
%e A192722 .5.|..1...250....3900....14400....14400
%e A192722 .6.|..1...922...32400...252000...648000...518400
%e A192722 ...
%e A192722 T(4,2) = 68:
%e A192722 There are 3 compositions of 4 into 2 parts, namely, 4 = 2 + 2 = 1 + 3 = 3 + 1; hence
%e A192722 T(4,2) = (4!/(2!*2!))^2 + (4!/(1!*3!))^2 + (4!/(3!*1!))^2
%e A192722 = 36 + 16 + 16 = 68.
%e A192722 Matrix identity: A192721 * Pascal's triangle = row reverse of A192722:
%e A192722 /...1................\ /..1..............\
%e A192722 |...3.....1...........||..1....1..........|
%e A192722 |..19....16.....5.....||..1....2....1.....|
%e A192722 |.211...299....65....1||..1....3....3....1|
%e A192722 |.....................||..................|
%e A192722 =
%e A192722 /...1...................\
%e A192722 |...4......1.............|
%e A192722 |..36.....18......1......|
%e A192722 |.576....432.....68.....1|
%e A192722 |........................|
%p A192722 J := unapply(BesselJ(0, 2*sqrt(-1)*sqrt(z)), z):
%p A192722 G := 1/(1-x*(J(z)-1)):
%p A192722 Gser := simplify(series(G, z = 0, 15)):
%p A192722 for n from 1 to 14 do
%p A192722 P[n] := n!^2*sort(coeff(Gser, z, n)) od:
%p A192722 for n from 1 to 14 do seq(coeff(P[n], x, k), k = 1..n) od;
%p A192722 # yields sequence in triangular form
%p A192722 # second Maple program:
%p A192722 b:= proc(n) option remember; expand(
%p A192722       `if`(n=0, 1, add(x*b(n-i)/i!^2, i=1..n)))
%p A192722     end:
%p A192722 T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n)*n!^2):
%p A192722 seq(T(n), n=1..14);  # _Alois P. Heinz_, Sep 10 2019
%t A192722 b[n_] := b[n] = Expand[If[n == 0, 1, Sum[x b[n-i]/i!^2, {i, 1, n}]]];
%t A192722 T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n] n!^2];
%t A192722 Table[T[n], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Dec 07 2019, after _Alois P. Heinz_ *)
%Y A192722 Cf. A001044, A002190, A061691, A192721, A102221 (row sums), A000275 (alternating row sums).
%K A192722 nonn,easy,tabl
%O A192722 1,3
%A A192722 _Peter Bala_, Jul 11 2011